\(y\cdot2+\frac{y}{2}=10\)
NT
Những câu hỏi liên quan
\(y=\frac{2^{19}\cdot27^3+15\cdot4^9\cdot9^4}{6^{9^{ }}\cdot2^{10}+12^{10}}\)tính y
Xem chi tiết
A=\(\frac{15\cdot3^{11}+4.27^4}{9^7}\)
B=\(\frac{2^{19}\cdot2^{73}+15\cdot4^9\cdot9^4}{6^9\cdot2^{10}+12^{10}}\)
C=\(\frac{5\cdot12^3\cdot4^{11}-16^8}{\left(3\cdot2^{17}\right)^2}\)
D=\(\frac{4^7\cdot2^8}{3\cdot2^{15}\cdot16^2-5\cdot2^2\cdot\left(2^{10}\right)^2}\)
1 ) Tìm x biết
a) \(x^{10}\cdot\left(x^2\right)^{10}\cdot\left(x^3\right)^{10}\cdot...\cdot\left(x^{10}\right)^{10}\)
b)\(\frac{1}{2}\cdot2^x+4\cdot2^x=9\cdot2^5\)
c)\(3\cdot2^{x+2}=5\cdot2^3\)
\(y=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{18\cdot19\cdot20}\)
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tham khảo nhé, vội quá ko trình bày cho bạn được
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Tính nhanh
C=\(\frac{4^7\cdot2^8}{3\cdot2^{15}\cdot16^2-5\cdot2^2\cdot\left(2^{10}\right)^2}\)
B=\(\frac{7\cdot6^{10}\cdot2^{20}\cdot3^6-2^{19}\cdot6^{15}}{9\cdot6^{19}\cdot2^9-4.3^{17}\cdot2^{26}}\)
\(\frac{1}{36}\)k cho mình nha
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Xem thêm câu trả lời
tìm y : (\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{8\cdot9\cdot10}\)) . y=\(\frac{23}{45}\)
Tinh
A=\(\frac{15\cdot3^{11}+4\cdot27^1}{9^7}\)
B=\(\frac{5\cdot2^{13}+4^{11}-1}{\left(3\cdot2^{17}\right)^2}\)
C=\(\frac{2^{19}\cdot27^3+15\cdot4^9\cdot9^4}{6^9\cdot2^{10}+12^{10}}\)
\(A=\frac{15.3^{11}+4.27^1}{9^7}\)
\(\Rightarrow A=\frac{3.5.3^{11}+4.3^{3^1}}{\left(3^2\right)^7}\)
\(\Rightarrow A=\frac{3^{12}.5+4.3^3}{3^{14}}\)
\(\Rightarrow A=\frac{3^3.\left(5.3^8+4.3^3\right)}{3^{14}}\)
\(\Rightarrow A=\frac{32805+4}{177147}\)
\(\Rightarrow A=\frac{32809}{177147}\)
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Tìm y :
\(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}\right)\cdot y=\frac{49}{200}\)
\(\Rightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right).y=\frac{49}{100}\)
\(\Leftrightarrow\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{100-98}{98.99.100}\right).y=\frac{49}{100}\)
\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right).y=\frac{49}{100}\)
\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{99.100}\right).y=\frac{49}{100}\Leftrightarrow\left(\frac{99.50-1}{99.100}\right).y=\frac{49}{100}\)
\(\Leftrightarrow\left(\frac{99.50-1}{99}\right).y=49\Leftrightarrow\left(99.50-1\right).y=99.49\Rightarrow y=\frac{99.49}{99.50-1}\)
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